1. The
channel capacity of a band-limited Gaussian
channel is given by:
(A) B log2(2+S/N)
(B) B log2(1+S/N)
(C) B log10(1+S/N)
(D) B loge(1+S/N)
Answer: B
Explanation:
C=B log2 (1 + S/N) where C is the capacity in bits per
second. B is the bandwidth of the channel in Hertz and S/N is the signal to
noise ratio.
2. The graph K3,4 has:
(A) 3 edges
(B) 4 edges
(C) 7 edges
(D) 12 edges
Answer: D
Explanation:
A bipartite graph is a complete bipartite graph if
every vertex in U is connected to every vertex in V. If U has n elements and V
has m, then the resulting complete bipartite graph can be denoted by Kn,m
and the number of edges is given by n*m. The number of edges Kn,m =
K 3,4
n*m = 3 * 4 =
12
3. The total number of spanning trees that can be drawn using five labeled
vertices is:
(A) 125
(B) 64
(C) 36
(D) 16
Answer: A
Explanation:
Cayley devised the well-known formula nn−2 for
the number of spanning trees in the complete graph Kn. = 55-2 = 125
4. Extremely low power dissipation and low cost per gate can be achieved
in:
(A) MOS ICs
(B) CMOS ICs
(C) TTL ICs
(D) ECL ICs
Answer: B
5. An example of a universal building block is:
(A) EX-OR gate
(B) AND gate
(C) OR gate
(D) NOR gate
Answer: D
Explanation:
Universal gates are the ones which can be used for
implementing any gate like AND, OR and NOT or any combination of these basic
gates. Apart from the NOR gate, NAND gate is also considered as universal gate.
6. An example of a layer that is absent in broadcast networks is:
(A) Physical layer
(B) Presentation layer
(C) Network layer
(D) Application layer
Answer: C
7. The ATM cell is:
(A) 48 bytes long
(B) 53 bytes long
(C) 64 bytes long
(D) 69 bytes long
Answer: B
Explanation:
An ATM cell always consists of a 5-byte header
followed by a 48-byte payload. So the size is 53 bytes long.
8. Four jobs J1, J2, J3 and J4
are waiting to be run. Their expected run times are 9, 6, 3 and 5 respectively.
In order to minimize average response time, the jobs should be run in the
order:
(A) J1 J2 J3 J4
(B) J4 J3 J2 J1
(C) J3 J4 J1 J2
(D) J3 J4 J2 J1
Answer: D
9. Suppose it takes 100 ns to access page table and 20 ns to access
associative memory. If the average access time is 28 ns, the corresponding hit
rate is:
(A) 100 percent
(B) 90 percent
(C) 80 percent
(D) 70 percent
Answer: B
Explanation:
h x 20 + (1-h) x 100 = 28
h=0.9 = 90%
10. Transmission of N signals, each band limited to fm Hz by TDM,
requires a minimum band-width of:
(A) fm
(B) 2 fm
(C) N fm
(D) 2N fm
Answer: C
Explanation:
Minimum transmission band-width of TDM channel is
given by the following equation. Bt=NW Where N is the total number of channels,
which are band limited to ‘W’ Hz.
In the above problem, the number of signals are N,
each band limited to Fm Hz.
So the minimum band-width =NW = N fm.
1 Comments
20 is c)
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